8. If a pipe whose diameter is 1.5 in. fill a cistern in 5 hours, In what time will a pipe whose diameter is 3 in. fill the same cistern? Ans. 1 hours. SUG.-It pours in 4 times as much, and fills it in 4 of 5 hours.

9. Required the distance between a lower corner and the opposite upper corner of a room 48 feet long, 20 feet wide, and 39 feet high.

644. There are Two Methods of explaining the general process of extracting the Cube Root, called the Analytic or Algebraic Method, and the Synthetic or Geometrical Method.

645. The Analytic Method of cube root is so called because it analyzes the number into its elements, and derives the process from the law of involution.

646. The Geometrical Method of cube root is so called because it makes use of a cube to explain the process.

1. Extract the cube root of 91125.

ANALYTIC SOLUTION.

-Since the cube of a number consists of three times as many places as the number itself, or of three times as many less one or two, the cube root of 91125 will consist of two places, or of tens and units, and the number 'tself will consist of tens + 3x tens2 X units + 3 tens units2 + units3. The greatest number of tens whose cube is contained in 91125 is 4 tens. Cubing the tens and subtracting, we have 27125, which equals 3 × tens2X units + 3x tens

(3ť2+3t × u + u2) ×u = 5425 × 5=27125

X units2 + units. Now, since 3x tens? X units is much greater than 3 Xtens X units2+ units3, 27125 consists principally of 3 times tens2 × units; hence, if we divide by 3 times tens2, we can ascertain the units

=

3 times tens equals 3 x 4024800; dividing by 4800, we are the units to be 5. We then find 3 times tens X units equal to 3 x 46 x 5600, and units252 25, and adding these and multiplying by unus we have (3 tens +3 tens units + units) X units, which equals 5425 x 5 =27125; subtracting, nothing remains, hence the cube root of 91125 is 45 Fig. 3.

Fig. 1.

R 40 6

40x52

Fig. 4.

GEOMETRICAL SOLUTION.-Let Fig. 1 represent the cube which contains 91125 cubic units, then our object is to find the number of linear units in its edge. The number of terms in the root, found as before, is two. The greatest number of tens whose cube is contained in the given number is 4 tens. Let A, Fig. 1, represent a cube whose sides are 40,

its contents will be 40364000. Subtracting 64000 from 91125 we find a remainder of 27125 cubic units, which by removing the cube A from Fig. 1, leaves a solid represented by Fig. 2.

Inspecting this solid, we perceive that the greater part of it consists of the three rectangular slabs, B, C, and D, each of which is 40 units in length and breadth, hence if we divide 27125 by the sum of the areas of one face of each regarded as a base, we can ascertain their thickness. The area of a face of one slab is 4021600, and of the three, 3 × 1600, 4800, and dividing 27125 by 4800 we have a quotient of 5, hence the thickness of the slab is 5 units.

Removing the rectangular slabs, there remain three other rectangulai solids, E, F, G, as shown in Fig. 3, each of which is 40 units long and 5 units thick, hence the surface of a face of each is 40 x 5200 square units. and of the three it is 3 x 40 x 5600 square units.

Finally removing E, F, and G, there remains only the little corner cube II, Fig. 4, whose sides are 5 units, and the surface of one of its faces 52-25 square units. We now take the sum of the surfaces of the solids remaining after the removal of the cube A, and multiply this by the common thickness, which is 5, and we have their solid contents equal to (4800+600+25) × 5 = 27125 cubic units, which, subtracted from the number of cubic units remaining after the removal of A, leaves no remainder. Hence the cube which contains 91125 cubic units is 40+5, or 45 units on a side.

NOTE. This can also be explained by building up the cube instead of separating it into its parts, for which see Manual.

647. We will now solve a problem with three figures in the root, indicating the solution by means of letters, and abbreviating the operation as in practice. A point like a period indicates the multiplication of the letters.

NOTES.-1. By the geometric method, when there are more than two fig ures we remove the first cube, rectangular slabs and solids, and small cube, and we have remaining three slabs, three solids, and a small cube, as before. 2. The method employed in actual practice is derived from the other by omitting ciphers, using parts of the number instead of the whole number each time we obtain a figure of the root, etc. It will also be seen that by separating the number into periods of 3 figures each, we have the number of places in the root, the part of the number used in obtaining each figure of the root, etc.

Rule.-I. Begin at units and separate the number into periods of three figures each.

II. Find the greatest number whose cube is contained in the left hand period, write it for the first term of the root, subtract its cube from the left hand period, and annex the next period to this remainder for a dividend.

III. Multiply the square of the first term of the root by 300 for 7 TRIAL DIVISOR; divide the dividend by it, and the result will be the second term of the root.

IV. To the trial divisor add 30 times the product of the

second term of the root by the first term, and also the square of the second term; their sum will be the TRUE DIVISOR.

V. Multiply the true divisor by the second term of the root subtract the product from the dividend, and annex the next period for another dividend. Square the root now found multiply by 300, and find the third figure as before, and thus continue until all the periods have been used.

NOTES.-1. If the product of the true divisor by the term of the root exceeds the dividend, the root must be diminished by a unit.

2. When a dividend will not contain a trial divisor, place a cipher in the root and two ciphers at the right of the trial divisor, bring down the next period, and proceed as before.

3. To find the cube root of a common fraction, extract the cube root of both terms. When these are not perfect cubes, reduce to a decimal and then extract the root.

4. By cubing 1, .1, .01, etc., we see that the cube of a decimal contains three times as many decimal places as the decimal; hence, to extract the cube root of a decimal, we point off the decimal in periods of three figures each, counting from the decimal point.

13=1

.13.001 .013.000001

SHORT METHOD OF CUBE ROOT.

C18. A Short Method of extracting the cube root is presented in the following modification of the ordinary method previously explained.

The abbreviation consists in obtaining the successive trial divisors by a law which enables us to use our previous work.

NOTE. This method was suggested to me by Dr. Geo. W. Hull, who has used it for several years with great satisfaction.

1. Extract the cube root of 14706125.

SOLUTION. We find as before the number of terms in the root and the first term of the root and cube, subtract and bring down the first period.

We then find as before the trial divisor, 12, by taking 3 times the square of the first term, and, dividing, find the second tern. of the root to be 4. We then as before take 3 times the product of the first and second terms and the square of the second term, and add these to the trial divisor as a correction to obtain the true divisor, 1456. We then multiply 1456 by 4, and subtract and bring down the next period.

We then, to find the next trial divisor, take the square of the last term, which is 16, and add it to the previous true divisor and the two corrections (which were added to the previous trial divisor), and we have 1728 as the next trial divisor.

Then, to find the true divisor, we add 3 times the product of the last term of the root into the previous part of the root, and also the square of the last term, and have 176425 for the true divisor. Multiplying by 5, we have 882125; hence the cube root is 245.

NOTE. No rule need be given, as the method is merely a modification of the previous method. The true divisor is obtained exactly as before; the abbreviation in obtaining the trial divisor is easily remembered by noticing that it equals the square of the last term, plus the true divisor, plus the correction used in finding that divisor.

The method is readily explained either by the blocks or by the algebraic formula. The method is indicated in the following formulas:

1. TRUE DIVISOR TRIAL DIVISOR+PRODUCT+SQUARE.
2. TRIAL DIVISOR=SQUARE+TRUE DIVISOR+CORRECTIONS.

2. Extract the cube root of 105154048.